Posts Tagged ‘dice’

The first is a pair of perfectly normal dice. The second is a “Merged d6”, a 36-sided die I bought through a crowdfunding campaign. Each of the sides is labeled with a sum of the results of rolling two normal dice. One of each of the even numbers between two and twelve is colored green. These let you simulate rolling doubles, as is required for games like Monopoly: each roll of two of the same number is represented. (The small print size of the numbers and the fact that it takes a moment to figure out which number is on top make this somewhat less practical than the normal dice, but I collect dice for mathematical interest rather than practicality.)

The blue and green dice are Sicherman dice. They are numbered a bit oddly. One of them has faces numbered 1, 2, 2, 3, 3, 4. The other’s faces are numbered 1, 3, 4, 5, 6, 8. The distribution of sums of die rolls is nevertheless the same as that of a normal pair of dice. In fact, it is the only non-standard numbering for a pair of dice with this property where the numbers are all positive integers.

Unfortunately, Sicherman dice don’t work for games that distinguish doubles rolls. Could we design a version of the Sicherman dice that does work for such games? The specially colored faces of the Merged d6 suggested a direction to take. We might color the faces of the Sicherman dice differently, and call a roll a doubles roll if the colors match. How many different ways are there to color the faces so as to produce exactly one match for every even number between two and twelve?

I posed this question on Google+. Joe DeVincentis found that there are sixteen essentially different solutions, of which two are the most interesting to me for designs that I might be interested in having made at some point:

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This has the advantage of only requiring three colors, and all of the faces that never match are on the same die. It also has an easy mnemonic that you could use even without coloring the faces: one on low die, odd on high die; four on low die, even on high die. Variations where some subset of the never-matching faces are a different color from the others are considered essentially similar here.

122334 — 134568

Here the colors can be ordered from low to high, and have the same order on each die.

The rest of the solutions follow. For clarity, I’ve colored all of the numbers that never match black even when they exist on both dice. We may define a rule that black is not considered to match itself.

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

122334 — 134568

One formulation of a magic cube simply numbers the vertices of a cube from one to eight, and requires that, for each face, the set of vertices that frame it have the same sum. There are three distinct magic cubes of this type:

Now, first of all, in applying this to a design of dice, it seems that the ideal application would be octahedral dice, since the octahedron is the dual of the cube, and thus the numbers could simply be printed on different faces. The sets of faces with magic sums would then be the ones surrounding a vertex of the octahedron. But I don’t have an order for custom d8’s, I have one for custom d6’s. Given that, what is the best way to represent one of these figures on a cube? As before, we can use a bold font to highlight the number that is used as the result of a roll. This time, since we can’t print directly on the vertices, the numbers will be repeated on each face that borders a vertex.

I have two answers. The first places the numbers on their traditional faces (with opposite faces summing to seven.) Only the middle numbering above can be used in this manner.

The second is suggested by the figure on the right. Since the numbers we aren’t using (that is, 7 and 8) appear at opposite vertices, we can highlight a diagonal ring around the cube and catch all of the numbers from 1 to 6. This is nicely symmetrical, but it does not put the numbers in their traditional positions.

A while back I supported a Kickstarter campaign for custom laser-engraved dice. I figured there had to be lots of mathy designs out there waiting to be found, and being able to make physical copies of them would inspire me to find some of them. And I have.

One of the first ideas i had was to use magic cubes. I didn’t know what sorts of magic cubes had been discovered, but I knew it was an obvious enough variation on the idea of magic squares that something had to be out there. In general, most magic cubes aren’t good for printing on dice, because they contain numbers in the interior of the cube, and one typically is only able to engrave the exterior. I did however find a couple of promising variations on the page on Unusual Magic Cubes at the late Harvey Heinz’s magic-squares.net site.

That page shows a cube found by Mirko Dobnik where each face is divided into a 2×2 grid. The faces sum to 50, and the rings around the cube sum to 100. In order to turn Dobnik’s cube into a usable six-sided die, I would need to find a solution that placed the numbers 1 to 6 on different faces of the cube, ideally on the same faces that they would occupy on a standard die. Then I could set these numbers in boldface to show that they represent the result of a die roll for a given side.

I decided this was a good time to learn how to use a constraint solver. I picked Numberjack because it uses Python, which is the language I am most comfortable with, and there was a magic square example that I could tweak. With face sum and ring sum constraints, and constraints to put the numbers 1 to 6 on the proper faces, (plus symmetry breaking constraints) I was getting at least hundreds of thousands of solutions. So I added constraints to make the four diagonals that traverse all six faces to sum to 75, and I fixed the positions of the numbers 1-6. The most symmetrical ways to pick corners of the faces of a die are to take all of the ones on one diagonal ring, or to take the corners that meet at antipodal vertices of the die. The first would violate the diagonal sum constraint, and the second bunched the relevant numbers up more than I liked, so I picked an arrangement that still has rotational symmetry about one of the axes through antipodal vertices, but that doesn’t have reflection symmetry. Then there are four ways to pick the axis of symmetry. At first I chose one at random, and came up with just eight solutions, one of which had odds and evens in a checkered pattern. Then I tried again with the axis going through the [1,2,3] and [4,5,6] vertices, and there were two parity-checkered solutions out of six, one of which is shown above.

I know it’s a bit strange to disappear from blogging for nearly a year, and then promise a multipart post that is itself part of a series, but yes, that is just what I’m doing. There are more magic dice to come, and then more mathematically inspired dice that are less magic. Hopefully there will also be some posts that are not about dice, not too far off.

I started by deciding that the pip positions should all connect North to south and East to West. It followed logically that I could have a puzzle where the solver could choose one of two possibilities for each empty cell: connecting North to West and South to East, or North to East and South to West. Because there are 33 non-pip squares, there would therefore be 233=8,589,934,592 ways to fill the grid. The lines on the outside of the grid show how the squares would connect when folded into a cube.

As an exercise, I found a way to make a single circuit, which is shown above. While that turned out to be about the difficulty of puzzle I can handle in something I am solving by hand, I’m sure there are more interesting specimens to be found.

Unfortunately, because the number of pips is odd, it’s impossible to have two circuits where each go through all of the pips. The circuits would have to cross each other an even number of times. But we could have one of the circuits cross itself once, and then have both circuits go through the remaining 20 pips. (Let’s call that problem number, oh what are we on, #12. By the way, the problem numbers are so that I can keep track of solvers of numbered problems and give them the fame they deserve. Nobody has solved any yet. You can be the first!)

Another possibility would be to use three circuits, each crossing itself once, and each visiting 12 pips in addition to the self-crossing. That’s #13.

I like big, wide ranging circuits here, so a constraint I like is to have circuits that visit all 6 sides. So bonus points on the preceding problem for having all three circuits do that. And that suggests #14: Maximize the number of circuits in a solution where all circuits present visit all 6 sides.

I thought early on that I could take this in the direction of a knot theoretical puzzle, but then one would have to keep track of which thread went under which in the crossings, which seemed like an unnecessary complication. I also think it would be interesting to make a multi-state maze (See Robert Abbott’s site for some good examples) using this template, but I haven’t yet had any good ideas for how that would work. If you have a good idea for a variation on this puzzle, I’d love to hear it. (This is of course true for all of my puzzles.)

For your solving convenience, I have an empty grid image here, and the Inkscape SVG file I used to produce it and the image above is here.

I have had a lifelong interest in recreational mathematics, especially polyomino problems. I’ve produced some puzzles in laser-cut plastic, which I sell via this very site. I’ve also dabbled in writing interactive fiction.